Problem: $ F = \left[\begin{array}{rr}4 & 1 \\ -2 & -2 \\ -1 & 4\end{array}\right]$ $ D = \left[\begin{array}{rr}4 & 2 \\ 1 & 4\end{array}\right]$ What is $ F D$ ?
Because $ F$ has dimensions $(3\times2)$ and $ D$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ F D = \left[\begin{array}{rr}{4} & {1} \\ {-2} & {-2} \\ \color{gray}{-1} & \color{gray}{4}\end{array}\right] \left[\begin{array}{rr}{4} & \color{#DF0030}{2} \\ {1} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{4}+{1}\cdot{1} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{4}+{1}\cdot{1} & ? \\ {-2}\cdot{4}+{-2}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{4}+{1}\cdot{1} & {4}\cdot\color{#DF0030}{2}+{1}\cdot\color{#DF0030}{4} \\ {-2}\cdot{4}+{-2}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{4}+{1}\cdot{1} & {4}\cdot\color{#DF0030}{2}+{1}\cdot\color{#DF0030}{4} \\ {-2}\cdot{4}+{-2}\cdot{1} & {-2}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{4} \\ \color{gray}{-1}\cdot{4}+\color{gray}{4}\cdot{1} & \color{gray}{-1}\cdot\color{#DF0030}{2}+\color{gray}{4}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}17 & 12 \\ -10 & -12 \\ 0 & 14\end{array}\right] $